Trig Cheat Sheet Definition of the Trig Functions Right triangle definition For this definition we assume that 0 2 p q < < or 0 90 q °< < ° . opposite sin hypotenuse q = hypotenuse csc opposite q = adjacent cos hypotenuse q = hypotenuse sec adjacent q = opposite tan adjacent q = adjacent cot opposite q = Unit circle definition For this definition q is any angle. sin 1 y y q = = 1 csc y q = cos 1 x x q = = 1 sec x q = tan y x q = cot x y q = Facts and Properties Domain The domain is all the values of q that can be plugged into the function. sin q , q can be any angle cos q , q can be any angle tan q , 1 , 0, 1, 2, 2 n n q p Ê ˆ π + = ± ± Á ˜ Ë ¯ … cscq , , 0, 1, 2, n n q p π = ± ± … secq , 1 , 0, 1, 2, 2 n n q p Ê ˆ π + = ± ± Á ˜ Ë ¯ … cot q , , 0, 1, 2, n n q p π = ± ± … Range The range is all possible values to get out of the function. 1 sin 1 q -£ £ csc 1 and csc 1 q q ≥ £- 1 cos 1 q -£ £ sec 1 and sec 1 q q ≥ £- tan q -• < < • cot q -• < < • Period The period of a function is the number, T, such that ( ) ( ) f T f q q + = . So, if w is a fixed number and q is any angle we have the following periods. ( ) sin wq Æ 2 T p w = ( ) cos wq Æ 2 T p w = ( ) tan wq Æ T p w = ( ) csc wq Æ 2 T p w = ( ) sec wq Æ 2 T p w = ( ) cot wq Æ T p w = q adjacent opposite hypotenuse x y ( ) , xy q x y 1 Formulas and Identities Tangent and Cotangent Identities sin cos tan cot cos sin q q q q q q = = Reciprocal Identities 1 1 csc sin sin csc 1 1 sec cos cos sec 1 1 cot tan tan cot q q q q q q q q q q q q = = = = = = Pythagorean Identities 2 2 2 2 2 2 sin cos 1 tan 1 sec 1 cot csc q q q q q q + = + = + = Even/Odd Formulas ( ) ( ) ( ) ( ) ( ) ( ) sin sin csc csc cos cos sec sec tan tan cot cot q q q q q q q q q q q q - = - - = - - = - = - = - - = - Periodic Formulas If n is an integer. ( ) ( ) ( ) ( ) ( ) ( ) sin 2 sin csc 2 csc cos 2 cos sec 2 sec tan tan cot cot n n n n n n q p q q p q q p q q p q q p q q p q + = + = + = + = + = + = Double Angle Formulas ( ) ( ) ( ) 2 2 2 2 2 sin 2 2sin cos cos 2 cos sin 2 cos 1 1 2sin 2 tan tan 2 1 tan q q q q q q q q q q q = = - = - = - = - Degrees to Radians Formulas If x is an angle in degrees and t is an angle in radians then 180 and 180 180 t x t t x x p p p = fi = = Half Angle Formulas ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 sin 1 cos 2 2 1 cos 1 cos 2 2 1 cos 2 tan 1 cos 2 q q q q q q q = - = + - = + Sum and Difference Formulas ( ) ( ) ( ) sin sin cos cos sin cos cos cos sin sin tan tan tan 1 tan tan a b a b a b a b a b a b a b a b a b ± = ± ± = ± ± = m m Product to Sum Formulas ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 sin sin cos cos 2 1 cos cos cos cos 2 1 sin cos sin sin 2 1 cos sin sin sin 2 a b a b a b a b a b a b a b a b a b a b a b a b = - - + È ˘ Î ˚ = - + + È ˘ Î ˚ = + + - È ˘ Î ˚ = + - - È ˘ Î ˚ Sum to Product Formulas sin sin 2sin cos 2 2 sin sin 2 cos sin 2 2 cos cos 2 cos cos 2 2 cos cos 2sin sin 2 2 a b a b a b a b a b a b a b a b a b a b a b a b + - Ê ˆ Ê ˆ + = Á ˜ Á ˜ Ë ¯ Ë ¯ + - Ê ˆ Ê ˆ - = Á ˜ Á ˜ Ë ¯ Ë ¯ + - Ê ˆ Ê ˆ + = Á ˜ Á ˜ Ë ¯ Ë ¯ + - Ê ˆ Ê ˆ - = - Á ˜ Á ˜ Ë ¯ Ë ¯ Cofunction Formulas sin cos cos sin 2 2 csc sec sec csc 2 2 tan cot cot tan 2 2 p p q q q q p p q q q q p p q q q q Ê ˆ Ê ˆ - = - = Á ˜ Á ˜ Ë ¯ Ë ¯ Ê ˆ Ê ˆ - = - = Á ˜ Á ˜ Ë ¯ Ë ¯ Ê ˆ Ê ˆ - = - = Á ˜ Á ˜ Ë ¯ Ë ¯
Definition of the Trig Functions Right triangle definition For this definition we assume that
02p
q< < or 0 90q° < < ° .
oppositesin
hypotenuseq = hypotenusecsc
oppositeq =
adjacentcoshypotenuse
q = hypotenusesecadjacent
q =
oppositetanadjacent
q = adjacentcotopposite
q =
Unit circle definition For this definition q is any angle.
sin
1y yq = = 1csc
yq =
cos1x xq = = 1sec
xq =
tan yx
q = cot xy
q =
Facts and Properties Domain The domain is all the values of q that can be plugged into the function. sinq , q can be any angle cosq , q can be any angle
tanq , 1 , 0, 1, 2,2
n nq pÊ ˆπ + = ± ±Á ˜Ë ¯
…
cscq , , 0, 1, 2,n nq pπ = ± ± …
secq , 1 , 0, 1, 2,2
n nq pÊ ˆπ + = ± ±Á ˜Ë ¯
…
cotq , , 0, 1, 2,n nq pπ = ± ± … Range The range is all possible values to get out of the function.
1 sin 1q- £ £ csc 1 and csc 1q q≥ £ - 1 cos 1q- £ £ sec 1 andsec 1q q≥ £ -
tanq-• < < • cotq-• < < •
Period The period of a function is the number, T, such that ( ) ( )f T fq q+ = . So, if w is a fixed number and q is any angle we have the following periods.
Trig Substitutions If the integral contains the following root use the given substitution and formula.
2 2 2 2 2sin and cos 1 sinaa b x x
bq q q- fi = = -
2 2 2 2 2sec and tan sec 1ab x a x
bq q q- fi = = -
2 2 2 2 2tan and sec 1 tanaa b x x
bq q q+ fi = = +
Partial Fractions
If integrating ( )( )
P xdx
Q x
ÛÙı
where the degree (largest exponent) of ( )P x is smaller than the
degree of ( )Q x then factor the denominator as completely as possible and find the partial fraction decomposition of the rational expression. Integrate the partial fraction decomposition (P.F.D.). For each factor in the denominator we get term(s) in the decomposition according to the following table.
Factor in ( )Q x Term in P.F.D Factor in ( )Q x Term in P.F.D
ax b+ A
ax b+ ( )k
ax b+ ( ) ( )1 2
2k
k
AA A
ax b ax b ax b
+ + ++ + +
L
2ax bx c+ + 2
Ax B
ax bx c
++ +
( )2 k
ax bx c+ + ( )1 1
2 2
k k
k
A x BA x B
ax bx c ax bx c
+++ +
+ + + +L
Products and (some) Quotients of Trig Functions
sin cosn mx x dxÚ
1. If n is odd. Strip one sine out and convert the remaining sines to cosines using 2 2sin 1 cosx x= - , then use the substitution cosu x=
2. If m is odd. Strip one cosine out and convert the remaining cosines to sines using 2 2cos 1 sinx x= - , then use the substitution sinu x=
3. If n and m are both odd. Use either 1. or 2. 4. If n and m are both even. Use double angle formula for sine and/or half angle
formulas to reduce the integral into a form that can be integrated. tan secn m
x x dxÚ 1. If n is odd. Strip one tangent and one secant out and convert the remaining
tangents to secants using 2 2tan sec 1x x= - , then use the substitution secu x= 2. If m is even. Strip two secants out and convert the remaining secants to tangents
using 2 2sec 1 tanx x= + , then use the substitution tanu x= 3. If n is odd and m is even. Use either 1. or 2. 4. If n is even and m is odd. Each integral will be dealt with differently.
Convert Example : ( ) ( )3 36 2 2cos cos 1 sinx x x= = -
A · (B£C) = B · (C£A) = (A£B) · CA£ (B£C) = B(A · C)°C(A · B)
f = f(r) g = g(r) A = A(r) B = B(r)
r(f + g) = rf +rg
r(fg) = f(rg) + g(rf)
r · (A + B) = r · A +r · Br · (fA) = f(r · A) + A · (rf)
r · (A£B) = B · (r£A)°A£ (r£B)
r£ (A + B) = r£A +r£B
r£ (fA) = f(r£A)°A£ (rf)
r · (r£A) = 0
r£ (rf) = 0
r£ (r£A) = r(r · A)°r2A
Di↵erential Equations Study Guide1
First Order Equations
General Form of ODE:dy
dx
= f(x, y)(1)
Initial Value Problem: y0 = f(x, y), y(x
0
) = y
0
(2)
Linear Equations
General Form: y0 + p(x)y = f(x)(3)
Integrating Factor: µ(x) = e
Rp(x)dx
(4)
=) d
dx
(µ(x)y) = µ(x)f(x)(5)
General Solution: y =
1
µ(x)
✓Zµ(x)f(x)dx+ C
◆(6)
Homeogeneous Equations
General Form: y0 = f(y/x)(7)
Substitution: y = zx(8)
=) y
0= z + xz
0(9)
The result is always separable in z:
(10)
dz
f(z)� z
=
dx
x
Bernoulli Equations
General Form: y0 + p(x)y = q(x)y
n(11)
Substitution: z = y
1�n(12)
The result is always linear in z:
(13) z
0+ (1� n)p(x)z = (1� n)q(x)
Exact Equations
General Form: M(x, y)dx+N(x, y)dy = 0(14)
Text for Exactness:@M
@y
=
@N
@x
(15)
Solution: � = C where(16)
M =
@�
@x
and N =
@�
@y
(17)
Method for Solving Exact Equations:
1. Let � =
RM(x, y)dx+ h(y)
2. Set
@�
@y
= N(x, y)
3. Simplify and solve for h(y).
4. Substitute the result for h(y) in the expression for � from step
1 and then set � = 0. This is the solution.
Alternatively:
1. Let � =
RN(x, y)dy + g(x)
2. Set
@�
@x
= M(x, y)
3. Simplify and solve for g(x).
4. Substitute the result for g(x) in the expression for � from step
1 and then set � = 0. This is the solution.
Integrating Factors
Case 1: If P (x, y) depends only on x, where
(18) P (x, y) =
My �Nx
N
=) µ(y) = e
RP (x)dx
then
(19) µ(x)M(x, y)dx+ µ(x)N(x, y)dy = 0
is exact.
Case 2: If Q(x, y) depends only on y, where
(20) Q(x, y) =
Nx �My
M
=) µ(y) = e
RQ(y)dy
Then
(21) µ(y)M(x, y)dx+ µ(y)N(x, y)dy = 0
is exact.
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Second Order Linear Equations
General Form of the Equation
General Form: a(t)y
00+ b(t)y
0+ c(t)y = g(t)(22)
Homogeneous: a(t)y
00+ b(t)y
0+ c(t) = 0(23)
Standard Form: y
00+ p(t)y
0+ q(t)y = f(t)(24)
The general solution of (22) or (24) is
(25) y = C
1
y
1
(t) + C
2
y
2
(t) + yp(t)
where y
1
(t) and y
2
(t) are linearly independent solutions of (23).
Linear Independence and The Wronskian
Two functions f(x) and g(x) are linearly dependent if there
exist numbers a and b, not both zero, such that af(x)+ bg(x) = 0
for all x. If no such numbers exist then they are linearly inde-pendent.
If y
1
and y
2
are two solutions of (23) then
Wronskian: W (t) = y
1
(t)y
02
(t)� y
01
(t)y
2
(t)(26)
Abel’s Formula: W (t) = Ce
�Rp(t)dt
(27)
and the following are all equivalent:
1. {y1
, y
2
} are linearly independent.
2. {y1
, y
2
} are a fundamental set of solutions.
3. W (y
1
, y
2
)(t
0
) 6= 0 at some point t
0
.
4. W (y
1
, y
2
)(t) 6= 0 for all t.
Initial Value Problem
(28)
8<
:
y
00+ p(t)y
0+ q(t)y = 0
y(t
0
) = y
0
y
0(t
0
) = y
1
Linear Equation: Constant Coe�cients
Homogeneous: ay
00+ by
0+ cy = 0(29)
Non-homogeneous: ay
00+ by
0+ cy = g(t)(30)
Characteristic Equation: ar
2
+ br + c = 0(31)
Quadratic Roots: r =
�b±pb
2 � 4ac
2a
(32)
The solution of (29) is given by:
Real Roots(r1
6= r
2
) : yH = C
1
e
r1t+ C
2
e
r2t(33)
Repeated(r1
= r
2
) : yH = (C
1
+ C
2
t)e
r1t(34)
Complex(r = ↵± i�) : yH = e
↵t(C
1
cos�t+ C
2
sin�t)(35)
The solution of (30) is y = yP + hH where yh is given by (33)
through (35) and yP is found by undetermined coe�cients or
reduction of order.
Heuristics for Undetermined Coe�cients(Trial and Error)
If f(t) = then guess that yP =
Pn(t) ts(A0 +A1t+ · · ·+Antn)
Pn(t)eat ts(A0 +A1t+ · · ·+Ant
n)eat
Pn(t)eatsin bt tseat[(A0 +A1t+ · · ·+Ant
n) cos bt
or Pn(t)eatcos bt +(A0 +A1t+ · · ·+Ant
n) sin bt]
Method of Reduction of Order
When solving (23), given y
1
, then y
2
can be found by solving
(36) y
1
y
02
� y
01
y
2
= Ce
�Rp(t)dt
The solution is given by
(37) y
2
= y
1
Ze
�Rp(x)dx
dx
y
1
(x)
2
Method of Variation of Parameters
If y
1
(t) and y
2
(t) are a fundamental set of solutions to (23) then
a particular solution to (24) is
(38) yP (t) = �y
1
(t)
Zy
2
(t)f(t)
W (t)
dt+ y
2
(t)
Zy
1
(t)f(t)
W (t)
dt
Cauchy-Euler Equation
ODE: ax2
y
00+ bxy
0+ cy = 0(39)
Auxilliary Equation: ar(r � 1) + br + c = 0(40)
The solutions of (39) depend on the roots of (40):
Real Roots: y = C
1
x
r1+ C
2
x
r2(41)
Repeated Root: y = C
1
x
r+ C
2
x
rlnx(42)
Complex: y = x
↵[C
1
cos(� lnx) + C
2
sin(� lnx)](43)
Series Solutions
(44) (x� x
0
)
2
y
00+ (x� x
0
)p(x)y
0+ q(x)y = 0
If x
0
is a regular point of (44) then
(45) y
1
(t) = (x� x
0
)
n1X
k=0
ak(x� xk)k
At a Regular Singular Point x
0
:
Indicial Equation: r2 + (p(0)� 1)r + q(0) = 0(46)
First Solution: y
1
= (x� x
0
)
r1
1X
k=0
ak(x� xk)k
(47)
Where r
1
is the larger real root if both roots of (46) are real or
